Lecture 2 - Stanford Universitysporadic.stanford.edu/quantum/lecture2.pdf · Lecture 2 Daniel Bump...

Sweedler Notation Bialgebras and monoidal categories Rigid categories

Lecture 2

Daniel Bump

May 23, 2019

V ∗ V

K V V ∗

K


Review of bialgebras and Hopf algebras

Recall that a bialgebra over a field K is a Vector space V withlinear maps

µ : H ⊗ H → H, ∆ : H → H ⊗ H,

η : K → H, ε : H → K ,

subject to the axioms. Associativity and coassociativity:

H ⊗ H ⊗ H H ⊗ H

H ⊗ H H

µ⊗1

1H⊗µ µ

µ

H H ⊗ H

H ⊗ H H ⊗ H ⊗ H

∆

∆ 1H⊗∆

∆⊗1H


Bialgebra axioms (continued)

Unit:H ⊗ H

H ⊗ K H

µ

∼=

1H⊗η

H ⊗ H

K ⊗ H H

µ

∼=

η⊗KH

Counit:

H ⊗ H

H ⊗ K H

1H⊗ε∼=

∆

H ⊗ H

K ⊗ H H

ε⊗IH∼=

∆


Bialgebra axioms (continued)

The augmentation and coaugmentation axioms:

H ⊗ H K × K

H K

ε×ε

µ ∼=

ε

K H

K × K H × H

η

∼= ∆

η×η

These say that the counit is an algebra homomorphism, andthat the unit is a coalgebra homomorphism, respectively.


Bialgebra axioms (concluded)

Let τ(a⊗ b) = b ⊗ a. Hopf:

H ⊗ H H ⊗ H ⊗ H ⊗ H H ⊗ H ⊗ H ⊗ H

H H ⊗ H

∆⊗∆

µ

1H⊗τ⊗1H

µ⊗µ

∆

The associativity and unit axioms: H is an algebra.The coassociativity and counit: H is an coalgebra.Augmentation axiom: counit is an algebra homomorphismDually, the unit is a coalgebra homomorphismThe Hopf axiom: ∆ is an algebra homomorphismEquivalently µ is a coalgebra homomorphism


Definition of a Hopf algebra

A Hopf algebra is a bialgebra with a linear map S : H → Hcalled the antipode satisfying

H H ⊗ H H ⊗ H

K H

∆

ε

1H⊗S

µ

η

H H ⊗ H H ⊗ H

K H

∆

ε

S⊗1H

µ

η

In the analogy between Hopf algebras and groups, thissubstitutes for g · g−1 = g−1 · g = 1.


Sweedler Notation (I)

We will use ordinary ring notation for the multiplication and unitµ and η. Thus if a,b ∈ H let a · b = µ(a⊗ b) and let 1H = η(1K ).Indeed we may identify K with a subring of the center of Husing η.

The Sweedler notation streamlines the formula

∆(a) =N∑

i=1

a′i ⊗ a′′i .

Instead, write∆(a) = a(1) ⊗ a(2).

We are omitting the summation from the notation.


Sweedler Notation (II)

Sweedler notation from the previous slide:

∆(a) = a(1) ⊗ a(2). (1)

The co-associativity

(∆⊗ 1H)∆(a) = (1⊗∆)∆(a)

means

a(1)(1) ⊗ a(1)(2) ⊗ a(2) = a(1) ⊗ a(2)(1) ⊗ a(2)(2).

We will writea(1) ⊗ a(2) ⊗ a(3)

for either of these decompositions. Note that a(2) has a differentmeaning here than it did in (1).


Axioms in Sweedler notation (I)

The counit axiom:

H ⊗ H

H ⊗ I H

1H⊗ε∼=

∆

H ⊗ H

I ⊗ H H

ε⊗IH∼=

∆

In Sweedler notation,

a = a(1)ε(a(2)) = ε(a(1))a(2).


Axioms in Sweedler notation (II)

The antipode axiom:

H H ⊗ H H ⊗ H

K H

∆

ε

1H⊗S

µ

η

H H ⊗ H H ⊗ H

K H

∆

ε

S⊗1H

µ

η

ε(a) = a(1)S(a(2)) = S(a(1))a(2).

We should writeηε(a) = · · ·

but we identify ε(a) ∈ K with its image under η.


Hopf axiom

If A and B are algebras, so is A⊗ B with multiplication

(a⊗ b)(a′ ⊗ b′) = (aa′)⊗ (bb′).

If µA, µB and µA⊗B are the multiplications in A, B and A⊗ B thismeans

µA⊗B = (µA ⊗ µB) ◦ (1A ⊗ τ ⊗ 1B).

Hence we write the Hopf axiom

H ⊗ H (H ⊗ H)⊗ (H ⊗ H)

H H ⊗ H

∆⊗∆

µH µH⊗H

∆

In other words, ∆ : H → H ⊗ H is an algebra homomorphism.


Hopf axiom in Sweedler notation

We may also translate the Hopf axiom into Sweedler notation.

H ⊗ H (H ⊗ H)⊗ (H ⊗ H)

H H ⊗ H

∆⊗∆

µH µH⊗H

∆

Applying this to x ⊗ y ∈ H ⊗ H gives

(xy)(1) ⊗ (xy)(2) = x(1)y(1) ⊗ x(2)y(2).


The antipode

PropositionIn a Hopf algebra S(ab) = S(b)S(a).

The proof of this will be good practice for Sweedler notation sowe will explain it carefully. This is an analog of the groupproperty (xy)−1 = y−1x−1 so the problem will be to translatethe proof of that fact into the Hopf algebra setting. Let usponder the middle expression in

(xy)−1 = (xy)−1xyy−1x−1 = y−1x−1.

The Hopf analog is

S(x(1)y(1))x(2)y(2)S(y(3))S(x(3))

where

x(1) ⊗ x(2) ⊗ x(3) = (∆⊗ 1)∆(x) = (1⊗∆)∆(x)


The antipode (continued)

The Hopf analog of (xy)−1xyy−1x−1 is

S(x(1)y(1))x(2)y(2)S(y(3))S(x(3)) (2)

where

x(1) ⊗ x(2) ⊗ x(3) = (∆⊗ 1)∆(x) = (1⊗∆)∆(x).

So imitating what we did for the group identity, we will unravelthis expression two ways, proving both

S(x(1)y(1))x(2)y(2)S(y(3))S(x(3)) = S(xy)

andS(x(1)y(1))x(2)y(2)S(y(3))S(x(3)) = S(y)S(x).

Using the antipode axiom y(1)S(y(2)) = ε(y)

S(x(1)y(1))x(2)y(2)S(y(3))S(x(3)) = S(x(1)y(1))x(2)ε(y(2))S(x(3)).

What just happened?


What just happened?

Here we are parsing

y(1) ⊗ y(2) ⊗ y(3) = (1⊗∆)∆(y) = y(1) ⊗ y(2)(1) ⊗ y(2)(2).

Now applying the map a⊗ b ⊗ c 7→ a⊗ bS(c) to both sides ofthis identity gives

y(1) ⊗ y(2)S(y(3)) = y(1) ⊗ S(y(2)(1))y(2)(2) = y(1) ⊗ ε(y(2)).

That is how we get

S(x(1)y(1))x(2)y(2)S(y(3))S(x(3)) = S(x(1)y(1))x(2)ε(y(2))S(x(3)).


Going back to (2), bear in mind that ε(a) is a scalar and can bemoved around at will in formulas. Also ε(a)ε(b) = ε(ab) sincethe counit is a ring homomorphism by the augmentation axiom.

S(x(1)y(1))x(2)y(2)S(y(3))S(x(3)) =S(x(1)y(1))x(2)ε(y(2))S(x(3)) =S(x(1)y(1))x(2)S(x(3))ε(y(2)) =S(x(1)y(1))ε(x(2))ε(y(2)) =S(x(1)y(1)ε(x(2)y(2))).

Now by the Hopf axiom

x(1)y(1) ⊗ x(2)y(2) = (xy)(1) ⊗ (xy)(2)

so

S(x(1)y(1))x(2)y(2)S(y(3))S(x(3)) = S((xy)(1)ε((xy)(2))) = S(xy).


The other side

Now let us unravel (1) a different way.

S(x(1)y(1))x(2)y(2)S(y(3))S(x(3)) =S(x(1)(1)y(1)(1))x(1)(2)y(2)(2)S(y(2))S(x(2)) = Hopf axiomS((x(1)y(1))(1))(x(1)y(1))(2)S(y(2))S(x(2)) =ε(x(1)y(1))S(y(2))S(x(2)) =S(ε(y(1))y(2))S(ε(x(1))x(2))

and so

S(x(1)y(1))x(2)y(2)S(y(3))S(x(3)) = S(y)S(x).


Slogans

Modules over a bialgebra are a monoidal categoryFinite-Dimensional Modules over a Hopf algebra are a rigidmonoidal categoryModules over a quasitriangular bialgebra are a braidedmonoidal categoryFinite-Dimensional Modules over a ribbon Hopf algebra area ribbon category

For the first item, if V and W are modules over H, then V ⊗Wis a module over H ⊗ H. Since ∆ is an H-modulehomomorphism H → H ⊗ H, we may use it to transport thismodule structure to H. In Sweedler notation

a(v ⊗ w) = a(1)v ⊗ a(2)w .

The coassociativity means that U ⊗ (V ⊗W ) and (U ⊗ V )⊗Whave the same module structure.


Bialgebras and monoidal categories

We have just seen that the category of modules over abialgebra is a monoidal category, and that the key to this is thecomultiplication.

There is a dual construction. Since the axioms of a bialgebraare self-dual, this statement has a dual one, that the categoryof comodules over a bialgebra is a monoidal category.

This fact arises in the theory of affine algebraic groups. If G isan affine group scheme over a field K , the affine algebra O(G)is a commutative Hopf algebra, and if V is a module for G, thenthe dual space of V is a comodule for O(G). For a proof, seeWaterhouse, Introduction to Affine Group Schemes Section 3.2.


Dual objects

Our second slogan is that modules over a Hopf algebra form arigid category. This is a category in which objects have duals.The archetype is the category finite-dimensional vector spacesover a field K .

Let V be an object in a monoidal category with unit object I. Wewill abstract the properties of a (left) dual. This comes withmorphisms evV : V ∗ ⊗ V → I and coevV : I → V ⊗ V ∗ calledevaluation and coevaluation.

The evaluation and coevaluation maps are subject to thefollowing axioms.

(1V⊗evV )◦(coevV ⊗1V ) = 1V , (evV ⊗1V∗)◦(1V∗⊗coevV ) = 1V∗ .


Example: finite-dimensional vector spaces

In the category of finite-dimensional vector spaces, I = KV ∗ = Hom(V ,K ) and evV (v∗ ⊗ v) = v∗(v), evaluating thelinear functional v∗ ∈ V ∗ at the vector v .

Continuing with the example of a finite-dimensional vectorspace, we have to define the coevaluation, which is now to be alinear map K → V ⊗ V ∗. We pick dual bases vi of V and v∗i ofV ∗ and define coevV (a) = a

∑vi ⊗ v∗i . You may check that this

is independent of V , and that the axioms are satisfied.


Diagrams for evaluation and coevaluation

We can represent evV : V ∗ ⊗ V → K and coevV : K → V ∗ ⊗ VThe diagram is to be read from top to bottom.

V ∗ V

K V V ∗

K

At the top of the diagram is a sequence of spaces which are tobe tensored together, hence V ∗ ⊗V in the left figure; so the twofigures represent morphisms V ∗ ⊗ V → K and K → V ⊗ V ∗


Diagrams for rigidity axioms (I)

The first axiom

(1V ⊗ evV ) ◦ (coevV ⊗1V ) = 1V

means that the following two diagram represents the identitymap V → V .

V K

K V

V V ∗ V

Here coevV ⊗1V : V = K ⊗ V → V ⊗ V ∗ ⊗ V and1V ⊗ evV : V ⊗ V ∗ ⊗ V → V ⊗ K = V .


Diagrams for rigidity axioms (II)

Identifying K ⊗ V = V ⊗ K in the above diagram, the K issuperfluous. Moreover we may rely on the reader to slice thediagram horizontally at intervals to obtain the sequence ofspaces V , V ⊗ V ∗ ⊗ V ; we have only to label the segments ofthe diagram.

V

V

V ∗

V

V

Then the axiom asserts that the above two morphisms areequal. (The second one is the identity map V → V .)


Diagrams for rigidity axioms (III)

Here is the dual axiom

(evV ⊗1V∗) ◦ (1V∗ ⊗ coevV ) = 1V∗ .

V ∗

V ∗

V

V ∗

V ∗

We see that the two axioms may be understood as assertingthat such bends may be straightened.


Uniqueness of the left dual

Suppose we have another left dual V ∗. Let evV and coevV bethe corresponding evaluation and coevaluation maps as Wecan construct morphisms φ : V ∗ → V ∗ and ψ : V ∗ → V ∗ by

φ = (evV ⊗IV∗)(1V∗ ⊗ coevV )

ψ = (evV ⊗ IV∗)(1V∗ ⊗ coevV )

V ∗

V ∗

V

V ∗

V ∗

V


Uniqueness of the dual (continued)

We may diagram the composition ψ ◦ φ:

V ∗

V

V

V ∗

V ∗

= V ∗

V

V

V ∗V ∗

V ∗

ψ ◦ φ = (evV ⊗IV∗)(1V∗ ⊗ coevV )(evV ⊗ IV∗)(1V∗ ⊗ coevV )

= (evV ⊗IV∗)(IV∗⊗IV⊗evV⊗IV∗)(IV∗⊗coevV⊗IV⊗IV∗)(IV∗⊗coevV )


Uniqueness of the dual (continued)

After switching the order of coevV and evV as above (just usingthe fact that ⊗ is a bifunctor) we may use the axioms twice toconclude that ψ ◦ φ = IV∗ :

V ∗

V

V ∗

=

V ∗

V ∗

Similarly φ ◦ ψ = IV∗ , so φ and ψ are inverse isomorphisms.This proves the uniqueness of the dual, and also illustrates theuse of diagrammatic methods in proofs.


Dual morphisms and rigid categories

We define an object V in a monoidal category to be rigid if ithas a left dual V ∗. The category is called rigid if every object is.

Suppose that V and W are objects in a rigid category andf : V →W is a morphism. Define a morphism f ∗ : W ∗ → V ∗ by

f ∗ = (evW ⊗1V∗)(1W∗ ⊗ f ⊗ 1V∗)(IW∗ ⊗ coevV ).

Representing f as a dot (left) then f ∗ is as on the right.

V

W V ∗

W ∗

W

V


Exercises

Exercise 1. Let f : V →W and g : W → U be morphisms in arigid category. Prove that (gf )∗ = f ∗g∗.

Let V be a vector space and V ∗ its dual. We will write 〈v∗, v〉instead of v∗(v) for the dual pairing.

Exercise 2. Let H be a Hopf algebra, V a finite-dimensionalmodule, and let V ∗ be its dual vector space. If a ∈ H andv∗ ∈ V ∗ define av∗ by

〈av∗, v〉 = 〈v∗,S(a)v〉.

Prove that this makes V ∗ into a module over H.

Exercise 3. Prove that the category of finite-dimensionalmodules over a Hopf algebra is a rigid category.

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